Is the Schrodinger equation covariant under Galilean transformations?

I am only asking this question so that I can write an answer myself with the content found here:


and here:


I learned about these pages in a comment to this answer by Ron Maimon. I think Ron Maimon is the original writer of this content.

This is creative commons, so it's ok to copy it here. It is not on any textbook on non-relativistic quantum mechanics that I know of, and I thought it would be more accessible (if to no one else, at least for myself) and safe here. I hope this type of question is not in disagreement with site policy.

  • $\begingroup$ Luc Vinet at University of Montreal. There were no lecture notes as this was a new course, but I think the redaction is going well. He is actually my current advisor, so I can ask him for an update, if you want. $\endgroup$ Commented Sep 11, 2016 at 0:56
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    $\begingroup$ This is discussed in the textbook Leslie E. Ballentine - Quantum Mechanics: A Modern Development; Section 4.3. $\endgroup$
    – ungerade
    Commented Sep 12, 2017 at 17:38
  • 1
    $\begingroup$ Related: Why the Galileo transformation are written like this in Quantum Mechanics? $\endgroup$
    – Qmechanic
    Commented Jun 26, 2018 at 16:05
  • $\begingroup$ Also discussed in text by Eugene Commins, Q.M. at pp.71-73. $\endgroup$
    – daniel
    Commented Jul 16, 2018 at 12:07
  • $\begingroup$ See also Landau & Lifshitz, QM, Vol. 3, p. 52 problem in $\S 17$. L&L consider the effect of Galilean transformation on plane wave solution. This seems relevant: scielo.org.mx/pdf/rmf/v63n2/0035-001X-rmf-63-02-185.pdf $\endgroup$
    – Qmechanic
    Commented Sep 13, 2022 at 15:47

4 Answers 4


Operator formalism

Galilean symmetry requires that $H(p)$ is quadratic in $p$ in both the classical and quantum Hamiltonian formalism. In order for Galilean boosts to produce a $p$-independent phase factor, $px - Ht$ must have a very special form -translations in $p$ need to be compensated by a shift in $H$. This is only true when $H$ is quadratic.

The infinitesimal generator of boosts in both the classical and quantum case is

$$ B = \sum_i m_i x_i(t) - t \sum_i p_i $$

where the sum is over the different particles, and $B$, $x$, and $p$ are vectors.

The Poisson bracket/commutator of $B\cdot v$ with $x$ and $p$ generate infinitesimal boosts, with $v$ the infinitesimal boost velocity vector:

$$ [B \cdot v, x_i] = vt $$

$$ [B \cdot v, p_i] = v m_i $$

Iterating these relations is simple since they add a constant amount at each step. By iterating, the $dv$s incrementally sum up to the finite quantity $V$:

$$ x \rightarrow x_i + Vt $$ $$ p \rightarrow p_i + m_i V $$

$B$ divided by the total mass is the current center of mass position minus the time times the center of mass velocity:

$$ B = M X_\text{cm} - t P_\text{cm} $$

In other words, $B/M$ is the current guess for the position that the center of mass had at time zero.

The statement that $B$ doesn't change with time is the center of mass theorem. For a Galilean invariant system, the center of mass moves with a constant velocity, and the total kinetic energy is the sum of the center of mass kinetic energy and the kinetic energy measured relative to the center of mass.

Since $B$ is explicitly time-dependent, $H$ does not commute with $B$, rather:

$$ \frac{dB}{dt} = [H, B] + \frac{\partial B}{\partial t} = 0 $$

This gives the transformation law for $H$ under infinitesimal boosts:

$$ [B \cdot v, H] = - P_\text{cm} v $$

The interpretation of this formula is that the change in $H$ under an infinitesimal boost is entirely given by the change of the center of mass kinetic energy, which is the dot product of the total momentum with the infinitesimal boost velocity.

The two quantities $(H, P)$ form a representation of the Galilean group with central charge $M$, where only $H$ and $P$ are classical functions on phase-space or quantum mechanical operators, while $M$ is a parameter. The transformation law for infinitesimal $v$:

$$ P' = P + Mv $$ $$ H' = H - P\dot{v} $$

can be iterated as before -$P$ goes from $P$ to $P + MV$ in infinitesimal increments of $v$, while $H$ changes at each step by an amount proportional to $P$, which changes linearly. The final value of $H$ is then changed by the value of $P$ halfway between the starting value and the ending value:

$$ H' = H - (P + \frac{MV}{2}) \cdot V = H - P \cdot V - \frac{MV^2}{2} $$

The factors proportional to the central charge $M$ are the extra wavefunction phases.

Boosts give too much information in the single-particle case since Galilean symmetry completely determines the motion of a single particle. Given a multiparticle time-dependent solution:

$$ \psi_t(x_1, x_2, ..., x_n) $$

with a potential that depends only on the relative positions of the particles, it can be used to generate the boosted solution:

$$ \psi'_t = \psi_t(x_1 + vt,...,x_n + vt) e^{i P_\text{cm} \cdot X_\text{cm} - \frac{M v_\text{cm}^2}{2}t} $$

For the standing wave problem, the motion of the center of mass just adds an overall phase. When solving for the energy levels of multiparticle systems, Galilean invariance allows the center of mass motion to be ignored.

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    $\begingroup$ FWIW: The Schrödinger equation is not only invariant under Galilean transformations. It is invariant under a larger group: the so-called Schrödinger Group. At some point you may want to add this into your answer, for completeness. $\endgroup$ Commented Jan 27, 2018 at 17:04

Free Schrödinger equation

Galilean boosts are transformations which look at the system from the point of view of an observer moving with a steady velocity $-v$. A boost must change the physical properties of a wavepacket in the same way as in classical mechanics:

$$ p' = p + mv $$

$$ x' = x + vt $$

so that the phase factor of a free Schrödinger plane wave:

$$ px - Et = (p' - mv)(x' - vt) - \frac{(p' - mv)^2}{2m} t = p'x' + E't - mvx' + \frac{mv^2}{2} t $$

is only different in the boosted coordinates by a phase which depends on $x'$ and $t$, but not on $p$.

An arbitrary superposition of plane wave solutions with different values of $p$ is the same superposition of boosted plane waves, up to an overall $x,t$ dependent phase factor. So any solution to the free Schrödinger equation $\psi_t(x)$ can be boosted into other solutions:

$$ \psi'_t(x) = \psi_t(x + vt) e^{imvx - i \frac{mv^2}{2}t} $$

Boosting a constant wavefunction produces a plane-wave. More generally, boosting a plane-wave:

$$ \psi_t(x) = e^{ipx - i \frac{p^2}{2m}t} $$

produces a boosted wave:

$$ \psi'_t(x) = e^{ip(x+vt) - i\frac{p^2}{2m}t + imvx - i\frac{mv^2}{2}t} = e^{i(p+mv)x + i\frac{(p+mv)^2}{2m}t} $$

Bossting the spreading Gaussian wavepacket:

$$ \psi_t(x) = \frac{1}{\sqrt{a + it/m}} e^{-\frac{x^2}{2a}} $$

produces the moving Gaussian:

$$ \psi'_t(x) = \frac{1}{\sqrt{a + it/m}} e^{-\frac{(x+vt)^2}{2a} + imvxx - i\frac{mv^2}{2}t} $$

which spreads in the same way.

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    $\begingroup$ I think that there might be a typo in your first line of algebra. Should the answer be $p'x'+E't-mvx-\frac{mv^{2}}{2}t$? $\endgroup$ Commented Sep 9, 2016 at 11:07

A free Schrödinger wave has the form $$\psi = N e^{i(\omega t - kx)} ~.$$ A Galilei transformation turns this into $$\psi = N e^{i(\omega t' - k(x'+vt'))} = e^{i((\omega -kv) t' - kx')} ~.$$ So $$\omega' = \omega - kv$$ and $$k'=k ~.$$ The conclusion is that the Schrödinger equation is not covariant under Galilei transformations.

The equation is covariant under the so-called Schrödinger group. However some operations in this group are not coordinate transformations, as they depend on particle mass. The wikipedia article of this group is opaque. See my arxiv.org submission https://arxiv.org/abs/quant-ph/0403013.


I would like to contribute with a similar answer, which follows Bargmann's original article on Projective Representations.

Bargmann's approach: In this context, we mean by $x,v,a$ vectors and by $R$ a rotation matrix. Consider the Schrödinger equation in three dimensions: \begin{equation} i \frac{\partial \psi(x,t)}{\partial t}+\frac{1}{2m} \Delta \psi(x,t)= V \psi(x,t). \end{equation} Let's look at the same state from another frame of reference, which is given by the Galilean transformations as follows: \begin{equation}\label{Galilei}\tag{1}x'^{j}= \sum_{k=1}^3 \limits R^{jk}x^k+v^jt+a^j;\end{equation} \begin{equation}\label{Galilei1}\tag{2} t'=t+b. \end{equation}

In the above transformations, $R$ are rotations, there are $3$ of them along each axis, $v$ represents Galilean boosts, which are also $3$, $a$ represents translations in space. Since we assume the space to be $3$-dimensional, there are 3 translations $a$. Finally, there is one translation in time, parametrized by $b$. Therefore, we have found that the Galilean group is a $10$-dimensional Lie group. In the following, we want our quantum theory to be invariant under this symmetry group, therefore we have to represent it on the Hilbert space. Two physical states are equivalent, if the corresponding wavefunctions are equal up to a phase. This means, that we look at projective representations of the Galilean group on the Hilbert space. This is also extremely important, because it tells us that if we want Galilean covariance to be satisfied we must have complex wave functions. It is not a choice of description as in electrodynamics. Writing this out explicitly, we have: \begin{equation}\tag{2}\label{phase} \psi(x,t)=e^{if(x',t')} \psi'(x',t'). \end{equation} where $(x',t')$ depend on $(x,t)$ according to \ref{Galilei}, \ref{Galilei1}. We know that $\psi$ satisfies the Schrödinger equation. We want to determine $f$ such that $\psi'(x',t')$ satisfies it too: \begin{equation}\tag{3}\label{schrodingertransformed} i \frac{\partial \psi(x,t)}{\partial t}+ \frac{1}{2m} \Delta \psi(x,t)= V \psi(x,t). \end{equation} Since $f(x',t')$ is yet unknown, this restriction, namely the covariance has to determine it. Let's see how we could rewrite the above equation in terms of $x',t'$. To do that, first consider the coordinate transformations. They imply the following change of partial derivatives: \begin{equation}\tag{4}\label{galileiderivative} \begin{aligned} \nabla_i= \frac{\partial}{\partial x^i}&=\sum_{j=1}^{3} \limits\underbrace{\frac{\partial x^{'j}}{\partial x^i}}_{=R^{ji}} \frac{ \partial}{\partial x^{'j}}+ \underbrace{\frac{ \partial t'}{\partial x^i}}_{=0} \frac{\partial}{\partial t'}=\sum_{j=1}^{3} \limits R_{ji} \frac{\partial}{\partial x^{'j}}=\sum_{j=1}^{3} \limits (R_{ij})^{T} \frac{\partial}{\partial x^{'j}}:=R^{-1} \nabla_i ' ;\\ \frac{\partial}{\partial t}&=\sum_{i=1}^{3}\underbrace{\frac{\partial x^{'i}}{\partial t}}_{=v^i} \frac{\partial}{\partial x^{'i}}+ \underbrace{\frac{\partial t'}{\partial t}}_{=1} \frac{\partial}{\partial t'}=\sum_{i=1}^{3} \limits v^i \frac{\partial}{\partial x{'i}}+ \frac{\partial}{\partial t'}:= v \cdot \nabla ' + \frac{\partial}{\partial t'}. \end{aligned} \end{equation} We also know that the Laplacian is rotationally invariant. Since the only part in the coordinate transformation for $x'$, which contains $x$ is the rotational part, we can conclude that the Laplacian is invariant under the Galilean transformations aswell. Therefore, we arrive at the following equation: \begin{equation} \left( i \frac{\partial}{\partial t'} +i v \cdot \nabla ' +\frac{1}{2m} \Delta ' - V\right) \left( e^{i f(x',t')} \psi'(x',t') \right). \end{equation} Before we do the product rule, let us recall a vector calculus identity, which will produce us an extra 'unintuitive' term: \begin{equation} \Delta(f_1 f_2)=(\Delta f_1) f_2+ f_1 (\Delta f_2)+2 (\nabla f_1) \cdot (\nabla f_2). \end{equation} To expand this, we have to use the product rule: \begin{equation*} i \underbrace{\frac{\partial}{\partial t'} \left(e^{if(x',t')} \right)}_{=i \frac{\partial{f(x',t')}}{\partial t'} e^{if(x',t')}} \psi'(x',t') +i e^{if(x',t')} \frac{\partial}{\partial t'} \psi'(x',t') + i\underbrace{(v \cdot \nabla') \left(e^{if(x',t')} \right)}_{=(iv \cdot \nabla ' f(x',t')) e^{if(x',t')}} \psi'(x',t')+ie^{if(x',t')} (v \cdot \nabla ')(\psi'(x',t')) \end{equation*} \begin{equation*} +\frac{1}{2m}\left( \underbrace{\Delta' e^{if(x',t')}}_{\text{term 1}} \right) \psi'(x',t') +\frac{1}{2m}e^{if(x',t')} ( \Delta' \psi'(x',t'))+\frac{2}{2m}\left(\underbrace{\nabla' e^{if(x',t')}}_{=(i \nabla' f(x',t'))e^{if(x',t')}} \right) \cdot (\nabla ' \psi'(x',t')) -V e^{if(x',t')} \psi'(x',t')=0. \end{equation*} With term 1 we still have plenty of work to do. Let's compute it explicitly: \begin{equation*} \Delta ' e^{if(x',t')}= \nabla' \cdot \left( \nabla ' e^{if(x',t')} \right)= \nabla' \cdot \left( e^{if(x',t')} (i \nabla ' f(x',t')) \right) \end{equation*} \begin{equation*} = e^{if(x',t')} \nabla'\cdot (i \nabla' f(x',t'))+ i \underbrace{\left(\nabla ' e^{if(x',t')} \right)}_{= (i \nabla' f(x',t'))e^{i f(x',t')}} \cdot \nabla' f(x',t') \end{equation*} \begin{equation*} =i \left(\nabla'^2 f(x',t') \right) e^{if(x',t')} +i \nabla ' f(x',t') (i \nabla' f(x',t'))e^{if(x',t')} \end{equation*} \begin{equation*} =\left(i \nabla^{'2} f(x',t')-\left(\nabla ' f(x',t') \right)^2 \right) e^{if(x',t')}. \end{equation*} So, we can simplify our expression after the product rule as follows: \begin{equation*} \begin{aligned} & \underbrace{-\frac{\partial f(x',t')}{\partial t'} e^{if(x',t')} \psi'(x',t')}_{1} + \underbrace{ie^{if(x',t')} \frac{\partial \psi'(x',t')}{\partial t'}}_{3}-\underbrace{v \cdot (\nabla' f(x',t'))e^{if(x',t')} \psi'(x',t')}_{1}+\underbrace{e^{if(x',t')}(iv \cdot \nabla ') (\psi'(x',t')}_{2}\\ & +\underbrace{\frac{1}{2m} \left(i \nabla'^2 f(x',t') - \left(\nabla' f(x',t')\right)^2 \right)e^{if(x',t')} \psi'(x',t')}_{1}+ \underbrace{\frac{1}{2m} e^{if(x',t')}\left(( \Delta' \psi'(x',t') \right) -V e^{-if(x',t')} \psi'(x',t')}_{3} \\ &+ \underbrace{\frac{1}{m}\left( i \nabla' f(x',t') \right) e^{if(x',t')} \cdot (\nabla ' \psi'(x',t'))}_{2}=0. \end{aligned} \end{equation*} Regrouping the terms in the groups $1,2,3$ yields: \begin{equation*} \left(\underbrace{-\frac{\partial f(x',t')}{\partial t'}+\frac{i}{2m} \nabla^{'2}f(x',t')- \frac{1}{2m}\left(\nabla' f(x',t') \right)^2 -v \cdot \nabla' f(x',t')}_{\text{condition 1}}\right) e^{i f(x',t')} \psi'(x',t') \end{equation*} \begin{equation*} + i\left(\underbrace{v +\frac{1}{m} \nabla' f(x',t')}_{\text{condition 2}} \right)e^{if(x',t')} \cdot \left( \nabla ' \psi'(x',t') \right) + e^{if(x',t')}\underbrace{\left( i\frac{ \partial}{\partial t'}+\frac{1}{2m} \Delta '-V \right) \psi'(x',t')}_{\text{Schrödinger}}=0. \end{equation*} The last part is the Schrödinger equation for $\psi'(x',t')$. So we want to kill all the other terms in order for this to be equal to 0 on the RHS to guarantee covariance. We get the following conditions on $f$ to assure covariance: \begin{equation}\label{conditions1} \tag{4} v+\frac{1}{m} \nabla ' f(x',t')=0 ; \end{equation}

\begin{equation}\label{conditions2} \tag{5} - \frac{\partial f(x',t')}{\partial t'}+\frac{i}{2m}\nabla^{'2}f(x',t')- \frac{1}{2m}\left(\nabla' f(x',t') \right)^2 -v \nabla' f(x',t')=0. \end{equation} The above system can be easily integrated to obtain: \begin{equation} f(x',t')=-mv \cdot x'+\frac{1}{2}mv^2 t' + C. \end{equation} To check that this is indeed the case, let's substitute back into \ref{conditions1},\ref{conditions2}. To do this, compute all the objects appearing in the conditions: \begin{equation} \nabla' f(x',t')=-mv; \end{equation} \begin{equation} \frac{\partial f(x',t')}{\partial t'}=\frac{1}{2}mv^2; \end{equation} \begin{equation} \nabla^{'2}f(x',t')=0. \end{equation} Therefore, we have: \begin{equation*} v+\frac{1}{m}(-mv)=v+(-v)=v-v=0. \end{equation*} \begin{equation*} -\frac{1}{2} mv^2+\underbrace{\frac{-i}{2m}0}_{=0}-\underbrace{\frac{1}{2m} m^2 v^2}_{=\frac{1}{2} mv^2} - \underbrace{v(-mv)}_{=mv^2}=0 \implies -\frac{1}{2} mv^2 -\frac{1}{2} mv^2 + mv^2=0 \implies 0=0. \end{equation*} This shows that the above function $f(x',t')$ solves the system. So the wavefunctions transform under a Galilean transformation as: \begin{equation} \psi(x,t)=e^{i f(x',t')}\psi'(x',t') \implies \psi'(x',t')=e^{-if(x',t')} \psi(x,t). \end{equation} Writing out $f(x',t')$ explicitly: \begin{equation}\label{representation} \psi'(x',t')=e^{i \left( mv \cdot x' - \frac{1}{2}mv^2t'+C \right)} \psi(x,t). \end{equation} Or equivalently: \begin{equation}\label{representation1} \psi'(Rx+vt+a,t+b)= e^{i \left(mv \cdot (Rx+vt+a)- \frac{1}{2} mv^2(t+b) +C \right)} \psi(x,t). \end{equation} Since we have derived the most general case, let's discuss the textbook examples. One widely known textbook example, for example the one discussed by Merzbacher in his Quantum Mechanics book is the Galilean boost invariance. To do that, we have to set $R=1,a=b=0$: \begin{equation} \psi'(x+vt,t)=e^{i \left( mv \cdot(x+vt) - \frac{1}{2} mv^2t \right)} \psi(x,t). \end{equation} Or explicitly as it appears in the book: \begin{equation} \psi'(x,t)=e^{im \left( v \cdot x - \frac{1}{2}v^2t \right)} \psi(x-vt,t). \end{equation}


  1. Valentine Bargmann, On Unitary Ray Representations of Continuous Groups, Annals of Mathematics, 1954.

  2. Eugen Merzbacher, Quantum Mechanics, Third Edition

Special thanks goes to Valter Moretti, who helped me spot a mistake in my computation back then when I was doing it and struggled to get the equivalence of the results.

  • 2
    $\begingroup$ Last formula is Merzbacher, QM, p. 75-76 eq. (4.111). $\endgroup$
    – Qmechanic
    Commented Sep 12, 2022 at 11:40

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